COURSE:
CE 201 (STATICS)
LECTURE NO.:
28 to 30
FACULTY:
DR. SHAMSHAD AHMAD
DEPARTMENT:
CIVIL ENGINEERING
UNIVERSITY:
KING FAHD UNIVERSITY OF PETROLEUM
& MINERALS, DHAHRAN, SAUDI ARABIA
TEXT BOOK:
ENGINEERING MECHANICS-STATICS
by R.C. HIBBELER, PRENTICE HALL
LECTURE NO. 28 to 30
FRAMES AND MACHINES
Objectives:
►
To analyze the forces acting on the members of
frames and machines composed of pin-connected
members
FRAMES AND MACHINES
Frames and machines are two common types of
structures which are often composed of pin-connected
multi-force members
► Frames are generally stationary and are used to
support loads
► Machines contain moving parts and are designed to
transmit and alter the effect of forces
► Following are the examples of frames and machines:
►
P2
P2
P3
C
P1
D
C
B
P1
M
B
E
A
G
D
A
RAx
RAy
RDy
RAx
RGx
RAy
RGy
STEPS FOR THE ANALYSIS OF FRAMES AND MACHINES
Frames and machines can be analyzed using the following steps:
I. Free-Body Diagram
► Draw the FBD of either the entire structure, or a portion of
the structure, or each of the members of the structure
► The choice should be made so that it leads to the most direct
solution of the problem
► Forces common to two members which are in
contact act
with equal magnitude but opposite sense on the respective
FBD of the members
II. Equations of Equilibrium
► Apply the equations of equilibrium (i.e., ΣFx = 0, ΣFy = 0, ΣMo
= 0) to each of the FBDs considered for the analysis of the
structure
► There will be one extra equation of equilibrium for each of the
hinge joints in the FBD (i.e., ΣM about the hinge joint = 0)
PROBLEM SOLVING: Example # 1
Determine the force P needed to hold the 20-lb
block in equilibrium.
PROBLEM SOLVING: Example # 1
The force P can be determined by
applying equilibrium conditions to
the above free-body diagrams, as
follows:
At C
Σ Fy = 0 ⇒ T – P – P = 0
⇒ T = 2P
(1)
At A
Free-body diagram of the pulleys at C Σ Fy = 0 ⇒ P + T + P – 20 = 0
⇒ T + 2P = 20
(2)
and A are shown below:
T
T
P
P
C
P
P
F.B.D. at C
A
20lb
F.B.D. at A
Substituting T = 2P in Eq. (2) from
Eq. (1)
4P = 20
⇒ P = 5 lb
Ans.
PROBLEM SOLVING: Example # 2
The compound beam is supported by a
rocker at B and is fixed to the wall at A. If
it is hinged (pinned) together at C,
determine the reactions at the supports.
PROBLEM SOLVING: Example # 2
⇒
1600 sin 60D + 4000
By =
12
= 448.80 lb
Free-body diagram of the portion CB
of the compound beam is shown
below:
200lb
Cx
C
Hinge
Cy
4000lb-ft
60o
8ft
Ans.
Free-body diagram of the entire
compound beam is shown below:
500lb
MA
13
5
Ax
Ay
4ft
200lb
12
4ft
4000lb-ft
60o
8ft
4ft
By
(F.B.D. of entire structure)
4ft
(F.B.D. of BC portion)
By
Applying equilibrium conditions to the
above free-body diagram, as follows:
Applying equilibrium conditions to the ΣFx = 0
5
above free-body diagram, as follows:
⇒ Ax – 500 × 13 + 200 cos 60° = 0
ΣM about C = 0
⇒ Ax = 92.307 lb Ans.
⇒ 4000 – 12 By + (200 sin 60°) × 8 = 0
PROBLEM SOLVING: Example # 2
500lb
MA
13
5
Ax
Ay
4ft
200lb
12
4000lb-ft
60o
4ft
8ft
4ft
By
(F.B.D. of entire structure)
Σ Fy = 0
⇒ Ay – 500 ×
12
13
ΣM about B = 0
12
⇒ 20 Ay – MA – 500 × 13 × 16 – 200
sin 60° + 4000 = 0
⇒ 20 Ay – MA = 4077.43
− 200 sin 60° + By = 0
⇒ Ay + By = 634.74 lb
Substituting the value of Ay calculated
Substituting the value of By calculated earlier,
earlier,
MA = 20 × 185.94 – 4077.43 = − 358.70
Ay = 634.74 − 448.80 = 185.94 lb Ans. = 358.70 lb-ft ( ) Ans.
PROBLEM SOLVING: Example # 3
Determine the horizontal and vertical
components of force that pins A and C
exert on the two member arch.
PROBLEM SOLVING: Example # 3
ΣM about A = 0
⇒ 2 × 0.5 + 1.5 × 1 – 3 Cy = 0
⇒ Cy = 0.833 kN Ans.
Σ Fy = 0
⇒ Ay + Cy – 2 = 0
⇒ Ay + Cy = 2
Substituting the value of Cy calculated
Free-body diagram of the entire arch earlier,
is shown below:
⇒ Ay = 2 – 0.833 = 1.167 kN Ans.
B
2kN
Σ Fx = 0
1.5kN
⇒ Ax – Cx + 1.5 = 0 ⇒Cx − Ax = 1.5
Free-body diagram of the portion AB
1m
A
C
of the arch is shown below:
x
x
0.5m
3m
2kN
Cy
Ay
(hinge)
B
Bx
(F.B.D. of entire structure)
By
1m
Applying equilibrium conditions to the
above free-body diagram, as follows:
1.5m
Ax
1.5m
Ay
(F.B.D. of portion AB)
PROBLEM SOLVING: Example # 3
(hinge)
B
Bx
2kN
By
1m
1.5m
Applying equilibrium conditions to the
above free-body diagram, as follows:
ΣMabout B = 0
⇒ −1.5 Ax + 1.5 Ay – 2 × 1 = 0
Ax
1.5m
⇒ Ax =
Ay
(F.B.D. of portion AB)
1.5 Ay − 2
1.5
=
1.5 × 1.167 − 2
1.5
= −0.166 kN
Ans.
Substituting the value of Ax in the
equation: Cx − Ax = 1.5, obtained
earlier
Cx = 1.5 + Ax
= 1.5 + (–0.166) = 1.333 kN
Ans.
PROBLEM SOLVING: Example # 4
Determine the support reactions
of the frame shown in the
following figure. Also calculate
the force in member GB.
PROBLEM SOLVING: Example # 4
Applying equilibrium conditions to the
entire frame:
ΣM about A = 0
⇒ RHy × 8 – 80 × 4 – 40 × 15 = 0
920
⇒ RHy = 8 = 115 kN Ans.
ΣFy = 0
⇒ 115 – 80 + RAy = 0
⇒ RAy = −35 kN = 35 kN (↓)
ΣFx = 0
⇒ 40 + RAx + RHx = 0
⇒ RAx + RHx = −40
Ans.
PROBLEM SOLVING: Example # 4
Applying equilibrium conditions to the
free-body diagram of the portion
CDE, as follows:
RCx
C
80kN
4m
D
E
REx
8m
RCy
ΣM about C = 0
REy
−8 × REy + 80 × 4 = 0
⇒ REy = 40 kN
PROBLEM SOLVING: Example # 4
REy = 40 kN
E
16m
J
REx
3
6m
8
FGB
G
9m
RHx
H
RHy = 115 kN
Applying equilibrium conditions to the
free-body diagram of the portion
EGH, as follows:
ΣM about J = 0
⇒−115 × 16 + 40 × 16 − RHx × 15 = 0
⇒ RHx = − 80 kN = 80 kN (←) Ans.
ΣFy = 0
3
115 + 8.54 FGB – 40 = 0
⇒ FGB = 213.6 kN Ans.
Substituting the value of RHx in
equation: RAx + RHx = −40, obtained
earlier:
RAx = −RHx −40
= −(−80) −40 = 40 kN (→)
Ans.
PROBLEM SOLVING: Example # 5
The two-member structure is connected at C by a
pin, which is fixed to BDE and passes through the
smooth slot in member AC. Determine the
horizontal and vertical components of reaction at
the supports.
PROBLEM SOLVING: Example # 5
Applying equilibrium conditions to
the above free-body diagram:
ΣM about A = 0
⇒ −5 C + 600 = 0
⇒ C = 120 lb
Free-body diagram of the portion
AC of the structure is as follows:
C
90-θ
4
3
θ
4
θ = tan − 1 = 53.13D
3
90 − θ = 36.87 D
600lb-ft
Ax
Ay
(F.B.D. of the portion AC)
5f t
tan θ =
C
ΣFx = 0
⇒ Ax – 120 cos (90 - θ) = 0
⇒ Ax = 120 × 0.8 = 96 lb
ΣFy = 0
⇒ Ay + 120 sin (90 - θ) = 0
⇒ Ay = −120 ×0.6 = − 72 lb
= 72 lb (↓) Ans.
Ans.
PROBLEM SOLVING: Example # 5
ΣFx = 0
⇒ −120 cos (90 − θ) – Ex = 0
⇒ Ex = 120 × 0.8 = 96 lb Ans.
ΣM about E = 0
Free-body diagram of the portion
BCDE of the structure is as follows:
500lb
ΣFy = 0
C=120
90-θ
B
Ex
E
C
3ft
⇒ −500 × 8 – 120 × 0.6 × 5 + 2Dy = 0
⇒ Dy = 2180 lb
Ans.
3ft
Dy
2ft
Ey
(F.B.D. of the portion B-C-D-E)
Applying equilibrium conditions to
the above free-body diagram:
⇒ Dy + Ey – 500 – 120 × 0.6 = 0
⇒ Dy + Ey = 572
⇒ Ey = 572 – 2180 = −1608 lb
= 1608 lb (↓)
Ans.
PROBLEM SOLVING: Example # 6
Analyze the frame
following figure:
shown
in
E
REx
2m
D
2m
G
H
C
4m
100 N
B
3m
6m
5m
RAx
A
RAy
2m
the
PROBLEM SOLVING: Example # 6
Applying equilibrium conditions to the
above free-body diagram:
E
REx
2m
ΣM about A = 0
D
⇒ −REx × 13 – 100 × 8 = 0
⇒ REx = −61.54 N = 61.54 N (←)Ans.
2m
G
H
C
4m
100 N
B
3m
6m
5m
RAx
A
RAy
2m
ΣFx = 0
⇒ RAx +REx = 0
⇒ RAx = – REx = – (–61.54)
= 61.54 N (→)
Ans.
ΣFy = 0
⇒ RAy – 100 = 0
⇒ RAy = 100 N (↑)
PROBLEM SOLVING: Example # 6
Free-body diagram of the portion GCH of
the structure is as follows:
E
REx
2m
D
2m
G
H
C
4m
100 N
B
3m
2m
6m
Applying equilibrium conditions to the
above free-body diagram:
5m
RAx
A
ΣM about C = 0 ⇒
RAy
Reactions at center, H, of the
pulley are shown below:
100 N
⇒ TGB = 250 N (T)
4
5
TGB × 3 – 100 × 6 = 0
3
ΣFx = 0 ⇒ 5 × 250 + RCx – 100 = 0
⇒ RCx = − 50 N = 50 N (←)
100 N
H
100 N
ΣFy = 0
100 N
Ans.
⇒−
4
5
Ans.
× 250 + RCy – 100 = 0
⇒ RCy = 300 N (↑)
Ans.
PROBLEM SOLVING: Example # 7
If each of the three uniform links of the
mechanism has a length L and weight W,
determine the angle θ for equilibrium. The
spring which always remains vertical, is unstretched when θ = 0°.
PROBLEM SOLVING: Example # 7
Free-body diagram of the portion
AB of the structure is as follows:
Ax
A
Ay
θ
Fs=
L/2
W
KL
Sin θ
2
L/2
(F.B.D. of portion AB)
Using the angle θ, the stretch s in
the spring can be determined as
follows:
A
L/2
L
s = 2 sinθ
s
FB
Applying equilibrium conditions to
the above free-body diagram:
ΣM about A = 0
⇒(W–
θ
B
KL
sin θ )
2
L
× 2 cos θ − FB × L cos θ = 0
⇒ FB = 0.5 W – 0.25 KL sinθ
(1)
PROBLEM SOLVING: Example # 7
FB = 0.5 W – 0.25 KL sinθ
(1)
Applying equilibrium conditions to
the above free-body diagram:
ΣM about D = 0
(FB + W) × L cos θ + W
⇒ FB + 1.5W = 0
L
× 2 cos θ
=0
(2)
Substituting FB from Eq. (1) in Eq. (2):
Free-body diagram of the portion
BCD of the structure is as follows: ⇒ 0.5 W – 0.25 KL sinθ +1.5 W = 0
⇒ 2 W = 0.25 KL sinθ
FB
Dx
D
B
W
θ
Dy
W
(F.B.D. of portion BCD)
C
8W
⇒ sinθ = KL
⇒
-1 ⎛ 8W ⎞
θ = sin
⎜
⎟
⎝ KL ⎠
Ans.
PROBLEM SOLVING: Example # 8
The structure is subjected to the loading shown. Member AD is
supported by a cable AB and roller at C and fits through a smooth
circular hole at D. Member ED is supported by a roller at D and a pole
that fits in a smooth snug circular hole at E. Determine the x, y, z
components of reaction at E and the tension in cable AB.
PROBLEM SOLVING: Example # 8
Coordinates of points A, B, C, and
G are as follows: A (0.6, 0.9, 0) m;
B (0, 0.9, 0.8) m; C (0, 0.9, 0) m;
and G (0.3, 0.9, 0) m
rAB = {− 0.6 i + 0.8 k}m
rAB = 0.36 + 0.64 = 1 m
rAB
∴FAB = r FAB =
AB
The tension in cable AB can be
determined by taking ΣM about C = 0
using the free-body diagram of the
portion AC, as follows:
C
FAB
G
A
F={-2.5k}kN
F={k
(F.B.D. of portion AC)
{ − 0.6 i + 0.8 k}
FAB
1
= FAB {- 6 i + 0.8 k} kN
rCA = {0.6 i}m and rCG = {0.3 i}m
ΣM about C = 0
⇒ rCA × FAB + rCG × F = 0
⇒ {0.6 i} × FAB {0.6 i + 0.8 k}
+ {0.3 i} × {- 2.5 k} = 0
⇒ −0.48 j FAB + 0.75 j = 0
⇒ FAB =
0.75
0.48
= 1.5625 kN Ans.
PROBLEM SOLVING: Example # 8
Ez k=0
MEz k=0
E
MEx i
MEy j
Ey j
D
Cxi
Ex j
Dzk
C
G
FAB
{-2-5k}
A
The x, y, z components of reactions
at E can be determined by
applying equilibrium conditions to
the free-body diagram of the entire
structure, as follows:
ΣF = 0
⇒ Ex i + Ey j + Dz k + Cx i
− 2.5 k – 0.9375 i + 1.25 k = 0
⇒ ( Ex + Cx−0.9375)i + (Ey)j
+ (Dz − 2.5 + 1.25) = 0
Equating the coefficients of the
unit vectors i, j, and k to zero,
Ex + Cx – 0.9375 = 0
(1)
Ey = 0
(2)
Dz – 2.5 + 1.25 = 0
(3)
⇒ Dz = 1.25 kN
PROBLEM SOLVING: Example # 8
Ez k=0
MEz k=0
E
MEx i
ΣMabout y axis = 0
MEy j
Ey j
D
Cxi
Ex j
Dzk
C
G
FAB
{-2-5k}
A
ΣM about z axis = 0
0.9 Cx – 0.9375 × 0.9 = 0
⇒ Cx = 0.9375
(4)
Substituting the value of Cx in Eq.
(1):
⇒ Ex = 0
Σ Mabout x axis = 0
⇒ MEx + Dz × 0.5 + 1.25 × 0.9
– 2.5 × 0.9 = 0
⇒ MEx = 1.25 × 0.9 – 1.25 × 0.5
= 0.5 kN-m
MEy – 1.25 × 0.6 + 2.5 × 0.3 = 0
⇒ MEy = 0
Summary:
⎧ FAB = 1.5625 KN
⎪E = 0
⎪ x
⎪Ey = 0
⎪⎪
Ans. ⎨ EZ = 0
⎪ M = 0.5 KN-m
⎪ Ex
⎪ M Ey = 0
⎪
⎪⎩ M Ez = 0
Multiple Choice Problems
1. The support reactions Ay and Dy of the frame shown below are
(a) 16.2 kN (↓) and 13.8 kN (↓)
(c) 15.0 kN (↓) and 15.0 kN (↓)
(b) 16.2 kN (↑) and 13.8 kN (↑)
(d) 15.0 kN (↑) and 15.0 kN (↑)
Ans: (a)
Feedback:
ΣM about D = 0
−130
= −16.2 kN = 16.2 kN ( ↓
8
∑ Fy = 0 ⇒ Ay + 30 + Dy = 0 ⇒ Dy = − Ay − 30 = −(−16.2) − 30 = −13.8 kN =13.8 kN ( ↓ )
8 Ay − 10 × 4 + 30 × (8 − 5) + 20 × 4 = 0 ⇒ 8 Ay = −130 ⇒ Ay =
Multiple Choice Problems
2. The support reactions Cx and Cy of the frame shown below are
(a) 577 kN (→) and 1000 N (↓)
(c) 500 kN (→) and 1000 N (↓)
(b) 577 N (←) and 1000 N (↑)
(d) 500 N (←) and 1000 N (↑)
Ans: (b)
Feedback:
Free-body diagram of the portion BC of the frame is shown below:
Applying equilibrium condition to the above free-body diagram,
∑ M = 0 ⇒ ( F sin 60 ) × 4 − 2000 × 2 = 0 ⇒ F = 1154.7 N
∑ F = 0 ⇒ F cos 60 − C = 0 ⇒ C = 1154.7 × cos 60 = 577 N ( ← )
∑ F = 0 ⇒ F sin 60 − 2000 + C = 0 ⇒ C = 2000 − 1154.7 × sin 60 = 1000 N ( ↑ )
D
about C
AB
AB
D
x
AB
y
AB
D
x
x
D
D
y
y